**Proof Through Geometry**

The figure shows an equilateral triangle ABC with DE parallel to BC and EF parallel to AB. D and G are on AB; F and H are on BC. GM and HN are perpendicular to AC. If AC = a, DG = d, FH = f, and MN = x prove that x = (a + d + f) / 2

Equilateral Triangle ABC

AC = a

DE // BC, EF // AB

DG = d

FH = f

GM perpendicular to AC

HN perpendicular to AC

**Given:**Equilateral Triangle ABC

AC = a

DE // BC, EF // AB

DG = d

FH = f

GM perpendicular to AC

HN perpendicular to AC

**Prove:**

x=(a+d+f)/2## Hint 1: DE // BC, EF // AB

By having parallel sides with our large equilateral triangle, ABC, we notice that triangle ADE and triangle EFC are also equilateral triangles. Therefore AD = DE = AE and

EF = FC = EC.

EF = FC = EC.

## Hint 2: Other Than a Triangle

Notice the parallelogram and use it to find more equal lines.

Therefore AE = DE = BF and EC = EF = BD.

Therefore AE = DE = BF and EC = EF = BD.

## Hint 3: Hint 3: Start with *a*

in *a+d+f*

Substitute a in for the line segments it represents from the triangle.

*a*= AE + EC## Hint 4: Create Midpoint *k *on *a*

There is now a line BK that splits the equilateral triangle in half.

We now have two 30-60-90 triangles. With these triangles come

special properties such as the hypotenus is twice as much as the base.

Therefore AB = 2AK and BC = 2KC

We now have two 30-60-90 triangles. With these triangles come

special properties such as the hypotenus is twice as much as the base.

Therefore AB = 2AK and BC = 2KC

## Hint 5: Mess With the *x*

Now

Therefore BG = 2GT and BH = 2VH. Use this information to substitute in the eaquarions and information gathered.

*x*= MK + KN = GT + VH and still holding the same properties as the previos hint.Therefore BG = 2GT and BH = 2VH. Use this information to substitute in the eaquarions and information gathered.